How would one find the maximal $n$ such that there exists an $n$subset $S$ of $\mathbb{Z}^+$ such that $\forall A\subseteq S, \sum_{a\in A}a$ is either a perfect square or a perfect cube, or can one go arbitrarily large. This is the case if one loosens the restrictions to any perfect power. Also, what about the slightly more general case for perfect squares, cubes, $...$ up to $k$th powers for some $k$.
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Dietmann and Elsholtz (Hilbert cubes in progressionfree sets and in the set of squares, Israel J. Math. 192 (2012), 59–66) have shown that if $S\subseteq[1, x]$ is a set such that all subset sums are contained in the set of squares, then $S\leq (8/e+o(1))(\log\log x)^2$, and if all subset sums are contained in the set of $k$th powers, $k\geq 3$, then $S\leq (4+o(1))\log\log x$. Their method is pretty flexible, so it should apply to your setting as well. 

